Answer
$\dfrac{\tan^5 x}{5}+c$
Work Step by Step
Consider the integral $\int \tan^4 x \sec^2 x dx$
The given integral can be re-written as:
$\int \tan^4 x \sec^2 x dx=\int \tan^4 x (\sec^2 x dx)$
Plug in $a= \tan x \implies da= \sec^2 x dx$
We use the formula: $\int x^n dx=\int\dfrac{x^{n+1}}{n+1} dx$
This implies that $\int a^4 da =\int \dfrac{a^5}{5}+C=\dfrac{\tan^5 x}{5}+c$
